i3.9. CHAPTER APPENDIX: ELIMINATING THE A 10 -CONFIGURATION 969that u; < v.;. Thus as v.; and u; are normal in CH(Ai)* = CH·(A1), it follows
that 02(L~*) S v.;. But this contradicts 13.8.34.3.
This final contradiction establishes Theorem 13.8.l.Observe in fact that Theorems 13.3.16, 13.6.1, and 13.8.1 complete the treat-
ment of Hypothesis 13.3. l for all possibilities for L / 02 ( L) (cf. 13.3.2.1) except
L3(2).13.9. Chapter appendix: Eliminating the A 10 -configuration
This section eliminates the shadow of the group A 10 , by ruling out the existence
of M E M with M ~ 84 wr Z2. We prove:
THEOREM 13.9.1. There is no simple QTKE-group G such that there exists
TE 8yl(G) and ME M(T) satisfying M ~ 84 wr Z2.
Throughout the section, we assume G, T, Mis a counterexample to Theorem
13.9.1. As usual we will begin with a number of preliminary lemmas describing the
structure of M.
Observe that J(M) = Mix M2 with Mi ~ 84, and MJ. = M2 for s an involution
in M - J(T). Let Ti := T n Mi and (t) := Z(Ti). Notice Z(T) = (z) is of order 2
where z := tt^8 • Let A:= 02(M), so that A~ E15. For X S: G, let Gx := Ca(X),
and set Gz := Gz/ (z).
Let G := A10 be the alternating group on n := {1, ... '10}, :;i,nd M the subgroup
of G permuting
{{1,2,3,4},{5,6,7,8},{9,10}}.
There is an isomorphism a : M-+ M such that T := a(T) E 8yl2(G) arid M E
M(T). Let Mi := a(Mi), z := a(z), etc. We may choose our isomorphism·a so
that Mi = G-5, ... ,io and z = (1, 2)(3, 4)(5, 6)(7, 8).
We will show that the 2-local subgroups and 2-fusion in G are the same as
that of G; this is a contradiction since G is quasithin while G is not. From time to
time, we use the identification a of M with M to compute facts about M and its
subgroup T.
LEMMA 13.9.2. (1) A(T) ={Ai: 1 Si S 4}, with Ai:= A, and B := A2 both
normal in T, while A3 = A4. Further J(T) =Tix T2 =AB and AnB = Z(J(T)).(2) Na(J(T)) = T.
{3) A and B are weakly closed in T with respect to G. Hence fusion in A iscontrolled by M = Na(A), and in B by Na(B).
(4) M = Na(A), a^0 n A= aM for each a EA, lzMI = 9, and ltMI = 6.
(5) t ~ zG.
{6) J(T) E 8yb(Gt)·
PROOF. Part (1) is an easy calculation. As M E M, M = Na(A). Let
X := Na(J(T)) and X* := X/J(T). Then X acts on A(T), and as M = Na(A),
T = NM(J(T)) = Nx(A), so J(T) is the kernel of the action of X on A(T). Thus
X s 8ym(A(T)) ~ 84 with Z2 ~ T E 8yb(X), so either X. = T, or X ~ 83.
The latter is impossible, as Aut(J(T)) is a 2-group. Thus (2) holds.
As J(T) is weakly closed in T, and each member of A(T) is normal in J(T), we
may apply the Burnside Fusion Lemma A.1.35 to these normal subsets to conclude
for each D E A(T) that v^0 n J(T) = nNa(J(T)), and hence v^0 n J(T) = DT